gd imagecopy

pspeakman

Ok wonderful people I am trying to use imagecopy to well you know copy part of an image onto another, it looks simple the function is easy to understand but it is not working, now i know my hosts support gd as i ahve used some other gd functions and it's not kicking up any errors when i run this, so if anyone can see what i'm doing wrong i would be eternally greatful

<?php
$myImage = imagecreatefromgif("main.gif");
$source = "hello.jpg";

imagecopy($myImage,$source, 30, 30, 0, 0, 20, 20);

imagegif($myImage,"mainnew.gif");

print "<html>
<head>
</head>
<body bgColor='cccccc'>
<center>
";
print "<br><img src='mainnew.gif'><br>";

?>

and all you see is the original main.gif.

Thanks Phil

ClarkF1

what version of the gd library does the server have???

From php.net:

Versions of GD older than gd-1.6 support GIF format images, and do not support PNG, where versions greater than gd-1.6 and less than gd-2.0.28 support PNG, not GIF. GIF support was re-enabled in gd-2.0.28.

You can echo the contents of gd_info() to find out.

pspeakman

I did an echo of gd_info() and it says Array.
I'll try and find out though, i'll try jpegs instead.
Thanks, Phil

ClarkF1

phpinfo() should work (you'll need to scroll down to find it)

The correct syntax is var_dump(gd_info());

samudasu

Your not using the functions correctly. You can't use a string value as the source for imagecopy().

$source = "hello.jpg"; // this is a string
should be
$source = imagecreatefromjpeg("hello.jpg"); // this is an image resource